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\(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3x\left(1-\dfrac{x-1}{x+1}\right)\)
\(\Rightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3x.\dfrac{x+1-\left(x-1\right)}{x+1}\)
\(\Rightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3x.\dfrac{2}{x+1}\)
\(\Rightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{6x}{x+1}\)
\(\Rightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}-\dfrac{6x}{x+1}=0\)
\(\Rightarrow\dfrac{\left(x+1\right)^2-\left(x-1\right)^2-6x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Rightarrow\dfrac{4x-6x^2+6x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Rightarrow\dfrac{10x-6x^2}{\left(x-1\right)\left(x+1\right)=0}\)
\(\Rightarrow10x-6x^2=0\)
\(\Rightarrow x-6x^2=0\)
\(\Rightarrow2x\left(5-3x\right)=0\)
\(\Rightarrow x\left(5-3x\right)=0\)
\(\Rightarrow5-3x=0\)
\(\Rightarrow3x=5\)
\(\Rightarrow x=\dfrac{5}{3}\)
a) \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3x\left(1-\dfrac{x-1}{x+1}\right)\)
\(\Rightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}3\left(1-\dfrac{x-1}{x+1}\right),\left(đk:x\ne1;x\ne-1\right)\)
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3-\dfrac{3\left(x-1\right)}{x+1}\)
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=3-\dfrac{3x-3}{x+1}\)
\(\Leftrightarrow\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{3x-3}{x+1}=3\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2-\left(x-1\right)^2+\left(x-1\right)\cdot\left(3x-3\right)}{\left(x-1\right)\left(x+1\right)}=3\)
\(\Leftrightarrow\dfrac{2\cdot2x+3x^2-3x-3x+3}{\left(x-1\right)\left(x+1\right)}=3\)
\(\Leftrightarrow\dfrac{4x+3x^2-3x-3x+3}{\left(x-1\right)\left(x+1\right)}=3\)
\(\Leftrightarrow\dfrac{-2x+3x^2+3}{\left(x-1\right)\left(x+1\right)}=3\)
\(\Leftrightarrow-2x+3x^2+3=3\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow-2x+3x^2+3=3\left(x^2-1\right)\)
\(\Leftrightarrow-2x+3x^2+3=3x^2-3\)
\(\Leftrightarrow-2x+3=-3\)
\(\Leftrightarrow-2x=-3-3\)
\(\Leftrightarrow-2x=-6\)
\(\Rightarrow x=3\left(đk:x\ne1,x\ne-1\right)\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
Mạnh dạn đưa pt 1 ẩn về 2 ẩn :)
Đặt \(\frac{x+3}{x-2}=u;\frac{x-3}{x+2}=v\)
Ta có:
\(u^2+6v=7uv\)
\(\Leftrightarrow\left(u-v\right)\left(u-6v\right)=0\)
Xét nốt nha!
Câu b là phân tích các kiểu ra dạng như thế này nhé !
\(\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
Hoặc là bạn dựa vào đó mà phân tích đến cái A là Ok
Câu 1a : tự kết luận nhé
\(2\left(x+3\right)=5x-4\Leftrightarrow2x+6=5x-4\Leftrightarrow-3x=-10\Leftrightarrow x=\frac{10}{3}\)
Câu 1b : \(\frac{1}{x-3}-\frac{2}{x+3}=\frac{5-2x}{x^2-9}\)ĐK : \(x\ne\pm3\)
\(\Leftrightarrow x+3-2x+6=5-2x\Leftrightarrow-x+9=5-2x\Leftrightarrow x=-4\)
c, \(\frac{x+1}{2}\ge\frac{2x-2}{3}\Leftrightarrow\frac{x+1}{2}-\frac{2x-2}{3}\ge0\)
\(\Leftrightarrow\frac{3x+3-4x+8}{6}\ge0\Rightarrow-x+11\ge0\Leftrightarrow x\le11\)vì 6 >= 0
1) 2(x + 3) = 5x - 4
<=> 2x + 6 = 5x - 4
<=> 3x = 10
<=> x = 10/3
Vậy x = 10/3 là nghiệm phương trình
b) ĐKXĐ : \(x\ne\pm3\)
\(\frac{1}{x-3}-\frac{2}{x+3}=\frac{5-2x}{x^2-9}\)
=> \(\frac{x+3-2\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{5-2x}{\left(x-3\right)\left(x+3\right)}\)
=> x + 3 - 2(x - 3) = 5 - 2x
<=> -x + 9 = 5 - 2x
<=> x = -4 (tm)
Vậy x = -4 là nghiệm phương trình
c) \(\frac{x+1}{2}\ge\frac{2x-2}{3}\)
<=> \(6.\frac{x+1}{2}\ge6.\frac{2x-2}{3}\)
<=> 3(x + 1) \(\ge\)2(2x - 2)
<=> 3x + 3 \(\ge\)4x - 4
<=> 7 \(\ge\)x
<=> x \(\le7\)
Vậy x \(\le\)7 là nghiệm của bất phương trình
Biểu diễn
-----------------------|-----------]|-/-/-/-/-/-/>
0 7
a) Ta có: -\(x^2\)+4x - 9
<=> - ( \(x^2\)- 4x + 4 ) - 5
<=> - ( x - 2 )\(^2\) - 5
Vì - ( x - 2 )\(^2\)\(\le\)0 <=> - ( x - 2 )\(^2\) - 5 \(\le\)-5 với mọi x
b) Ta có x\(^2\)- 2x + 9
<=> ( x\(^2\) - 2x +1 ) + 8
<=> ( x - 1 ) \(^2\)+ 8
Vì ( x - 1 ) \(^2\)\(\ge\) 0 <=> ( x - 1 ) \(^2\)+ 8 \(\ge\) 8 với mọi thực x
a,Ta có:\(-x^2+4x-9\)
\(\Leftrightarrow-\left(x^2-4x+4\right)-5\)
\(\Leftrightarrow-\left(x-2\right)^2-5\)
Vì \(-\left(x-2\right)^2\le0\Leftrightarrow-\left(x-2\right)^2-5\le-5\forall x\)
b.Ta có:\(x^2-2x+9\)
\(\Leftrightarrow\left(x^2-2x+1\right)+8\)
\(\Leftrightarrow\left(x-1\right)^2+8\)
Vì \(\left(x-1\right)^2\ge0\Leftrightarrow\left(x-1\right)^2+8\ge8\forall x\)
Thay x+y+z=1 vào biểu thức C, ta được:
\(C=\left(x+y+z-x\right)\left(x+y+z-y\right)\left(x+y+z-z\right)\)
\(C=\left(y+z\right)\left(z+x\right)\left(x+y\right)=\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
Ta có: \(x^3+y^3+z^3=\frac{1}{9}\Leftrightarrow\left(x+y+z\right)^3-3\left(x+y\right)\left(y+z\right)\left(z+x\right)=\frac{1}{9}\)
Thay x+y+z=1. Suy ra \(1-3\left(x+y\right)\left(y+z\right)\left(z+x\right)=\frac{1}{9}\)
\(\Leftrightarrow3\left(x+y\right)\left(y+z\right)\left(z+x\right)=\frac{8}{9}\)
\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=\frac{8}{9.3}=\frac{8}{27}\)
\(\Rightarrow C=\left(x+y\right)\left(y+z\right)\left(z+x\right)=\frac{8}{27}.\)
ĐS:...
a)
\(\dfrac{x-3}{5}+\dfrac{1-2x}{3}=6\\ < =>3x-9+5-10x=90\)
\(< =>3x-10x=90+9-5\\ < =>-7x=94\\ < =>x=-\dfrac{94}{7}\)
b)
\(\left(2x-3\right)\left(x^2+1\right)=0\\ < =>\left[{}\begin{matrix}2x-3=0\\x^2+1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x^2=-1\left(voli\right)\end{matrix}\right.\\ < =>x=\dfrac{3}{2}\)
c)
\(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\left(x\ne-1;x\ne2\right)\)
suy ra: \(2\left(x-2\right)-x-1=3x-11\)
\(< =>2x-4-x-1-3x+11=0\)
\(< =>2x-x-3x=4+1-11\\ < =>-2x=-6\\ < =>x=3\left(tm\right)\)
a) \(\dfrac{x-3}{5}+\dfrac{1-2x}{3}=6\)
\(\Leftrightarrow3\left(x-3\right)+5\left(1-2x\right)=90\)
\(\Leftrightarrow-4-7x=90\)
\(\Leftrightarrow x=-\dfrac{94}{7}\)
b) \(\left(2x-3\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow2x-3=0\) (Vì \(x^2+1>0\))
\(\Leftrightarrow x=\dfrac{3}{2}\)
c) \(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\left(Đk:x\ne-1;x\ne2\right)\)
\(\Leftrightarrow2\left(x-2\right)-\left(x+1\right)=3x-11\)
\(\Leftrightarrow x-5=3x-11\)
\(\Leftrightarrow x=3\)
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vãi banana ra số to vl=))